# There is a reason sudoku uses squares

There exists a 9x9 grid with the cells in one single row numbered 1-9 in order. The cells in the other 8 rows are initially empty. Note: The cells initially containing numbers can be in any one row; not necessarily the first.

Draw borders to divide the square into 9 non-intersecting continuous regions containing 9 cells each such that you make a sudoku-like puzzle with a unique solution. You may not add any additional numbers or hints. A sudoku-like puzzle, for this question, has the following rules:

• each cell in the 9x9 grid contains exactly 1 integer between 1 and 9 inclusive

• each row and column and bordered region contains each integer between 1 and 9 inclusive exactly once

• @Will I am asking the user to make a jigsaw sudoku such that all of the initial 9 numbers are in a single row. Jun 13, 2016 at 18:27
• An image would really help here. It sounds like you're saying the first column is all $1$s, the second column is all $2$s, etc. ("...cells in a single row numbered 1-9 in order"). But then you say "...each ... column ... contains each integer between 1 and 9...". That seems to be contradictory. Jun 13, 2016 at 18:45
• Something like this? i.imgur.com/sWdGcVX.png Jun 13, 2016 at 18:59
• I might be looking at this wrong, but can't you just use normal sudoku boundaries to solve this? Jun 13, 2016 at 19:08
• If you make this into a normal sudoku board, it will not have a unique solution. No Sudoku board with a unique solution has every been found with less than 17 squares filled in. youtube.com/watch?v=MlyTq-xVkQE Jun 13, 2016 at 19:16

There is a unique solution to the following

The solution is

Proof of uniqueness

Let us use the following notation:
The rows from top to bottom are given the labels $A-I$
The columns from left to right are numbered $1-9$ (as with the top row).

Firstly, $B9$ must be $1$ since it is the last in its continuous region. Then, this forces $C8$ to be $1$ since none of the rest of row $B$ can contain $1$ nor can column $9$. Similarly, we find that going down diagonally to the left all the entries are $1$ down to $I2$.

Now, look at $C9$. The entry here, $x$, must be the same as $D8$, since its continuous region has to contain $x$ but row $C$ and column $9$ already contain $x$. By a similar line of reasoning, we find, recursively, that the entries $E7$, $F6$, $G5$, $H4$ and $I3$ are all $x$ but of course cannot be $3,4,\ldots,9$ so $x=2$ and $B1$ must also be $2$.

We can continue this line of reasoning, next starting at the entry in $D9$, calling this $y$ and proceeding diagonally left and down to find $y=3$.

In this way, we can fill the entire grid, recursively always beginning at the topmost entry in column $9$.

Side notes and footnotes

The Sudoku variation in question turns out to be called “Du-Sum-Oh,” along with some aliases, and cells 1– 8 by themselves can force a unique solution without being given cell 9.

Hexomino’s original answer1 revealed how delightful this puzzle is but I had forgotten the details months later when mentioning it to a fellow Sudoku enthusiast, so some variety ensued. (Click within a spoiler to reveal it permanently.)

The layout on the left, with straightforward numbering, has a very sleek route to solution 2 whereas the numbering on the right demonstrates that an irregular set of initial numbers can also force a unique solution and be amusing to solve 3 if you’re in the mood.

Progress came from starting with small boards while experimenting with simple zigzags and L shapes. The 4×4 and 5×5 layouts along the way were misleadingly efficient 4 and led to an unnecessarily awkward 9×9 layout. Footnotes (solutions of layouts):

1 Synopsis of the three stages in Hexomino’s original solution. (Circles ◯ spotlight cells that were most recently filled or are immediately determinable at the steps shown.) 2 First and last steps of the present straightforward solution. (Circles ◯ mean the same1 as above.) 3 Synopsis of a solution for irregularly placed initial numbers. (Circles mean the same1 as above.) 4 Solutions of the 4×4 layout in just two steps and of the 5×5 layout in four steps. Edit: Angel Koh came up with another solution to my layout, so it is non-unique.

I believe to have a unique solution in regular sudoku you need at minimum: 1 number in each column, 1 number in each row, 1 number in each box, and every number from 1-9. But, you can cheat on 1 of these and for instance satisfy the remaining 3 clues but have a number in 8 boxes. Has a unique solution which is: Although I am not sure how to prove it.

• wouldn't swopping row 2 with any other rows still work? Jun 14, 2016 at 1:41
• @AngelKoh, ya I think you are right. Jun 14, 2016 at 1:47
• No, this solution is not unique. You can swap the 3rd and 4th line with the 5th and 6th line, for instance, and you have found annother solution. Mar 26, 2017 at 9:29
• @AngelKoh Swapping row 2 (from the top) with any of the following lines would invalidate at least the two blocks in the upper left and right. Mar 26, 2017 at 9:52